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svp [43]
2 years ago
15

How many bananas are equal to 7.50 moles of bananas?​

Chemistry
1 answer:
Veseljchak [2.6K]2 years ago
7 0

Answer:

4.52×10^24

Explanation:

N = n × Na

where; N = no. of bananas

n = no. of moles

Na = Avogadro's constant

Which is 6.02×10^23

N = 7.5 × 6.02×10^23

N =4.515×10^24

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So the GFM of O2 is 32.00g/mol
If we have 1 mole of O2 we have 32.00g
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How many grams of solute are present in 1250 mL of a 1.34 M NaNO3 solution?
inysia [295]
He answer is 42 because you carry the one
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The ph of a 0.30 m solution of a weak acid is 2.67. what is the ka for this acid?
azamat
To determine the Ka of the acid, we can use the equation for the pH of weak acids which is expressed as:

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7 0
3 years ago
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The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)
EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

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2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

3 0
3 years ago
If 36.2g of Acetic Acid (HC2H302) was dissolved in 300. mL of water, what is the
uysha [10]

Answer:

2.01 M

Explanation:

Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)

The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.

36.2g \times \frac{1mol}{60.05g} = 0.603mol

Step 2: Convert the volume of solution to liters

We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)

300mL \times \frac{1L}{1000mL} = 0.300L

Step 3: Calculate the molarity of the solution

The molarity is equal to the moles of solute (acetic acid) divided by the liters of solution

M = \frac{0.603mol}{0.300L} = 2.01 M

7 0
3 years ago
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